# The Multiplicative Structure of $$\Bbb Z / n \Bbb Z$$

$$\newcommand{\ZZ}{\Bbb Z} \newcommand{\ZZn}[1]{\ZZ / {#1} \ZZ}$$

One of the most familiar rings is the ring of integers modulo $$n$$, often denoted $$\ZZn{n}$$. Like all rings, it has an additive structure and a multiplicative one. The additive structure is straightforward: $$\ZZn{n}$$ is cyclic, generated by $$1$$. In fact, every integer $$a$$ coprime to $$n$$ is a generator for this group, giving a total of $$\phi(n)$$ generators. The multiplicative structure, on the other hand, is far less apparent.

Not all elements of $$\ZZn{n}$$ can participate in the multiplicative group, because not all of them have inverses. For example, 4 has no inverse in $$\ZZn{6}$$; there’s no integer $$a$$ such that $$4a \equiv 1 \pmod 6$$. Elements that do have inverses are called units, and we’ll denote the group of units in $$\ZZn{n}$$ as $$U_n$$.

Since an element $$a \in \ZZn{n}$$ is a unit iff $$a$$ and $$n$$ are coprime, there are $$\phi(n)$$ units, where $$\phi$$ is the totient function. But the size of the group alone doesn’t nail down the group structure.

For example:

• $$U_5 = \{ 1, 2, 3, 4 \}$$:
• generated by $$2$$: $$2^0 = 1$$, $$2^1 = 2$$, $$2^2 = 4$$, $$2^3 = 8 = 3$$
• also generated by $$3$$: $$3^0 = 1$$, $$3^1 = 3$$, $$3^2 = 4$$, $$3^3 = 2$$
• this group is isomorphic to $$\ZZn{4}$$
• $$U_8 = \{ 1, 3, 5, 7 \}$$
• every element squares to $$1$$
• this group is isomorphic to $$\ZZn{2} \times \ZZn{2}$$

Is there a way to find the structure of $$U_n$$?